Multiplying or adding constants within $P(X \leq x)$? Is it always true, that, for a positive constant $c$, we have $P(X \leq x) = P(c X \leq cx)$ for some continuous random variable $X$?
Further is it always true that $P(X + c \leq x + c) = P(X \leq x)$?
If so, what governs the equality?
 A: When you write $P(X \leq x)$, hidden in the background (if you haven't learned measure-theoretic probability) is the following:

*

*$X: \Omega \to \mathbb{R}$ is a real-valued random variable.

*$P$ is a probability measure.

The actual meaning of $P(X \leq x)$ is:
$$P(X \leq x) = P(\{\omega \in \Omega: X(\omega)\leq x\})\text{.}$$
This is regardless of whether $X$ is a continuous or discrete random variable.
The following facts are necessary to show that your equalities are true:

*

*Fact. If $X$ is a random variable, so is $Y = X + c$ for some constant $c \in \mathbb{R}$.

*Fact. If $X$ is a random variable, so is $Z = cX$ for some constant $c \in \mathbb{R}$.

Hence, we may see that
$$\begin{align}
P(Y \leq x + c) &= P(\{\omega: Y(\omega) \leq x + c\}) \\
&= P(\{\omega: (X + c)(\omega) \leq x + c\}) \\
&= P(\{\omega: X(\omega) +c\leq x + c\})
\end{align}$$
which is clearly equal to $P(X \leq x)$.
One may show similarly for $P(Z \leq cx)$ for some $c > 0$.
A: You can do all those things.
$X \le x$ is an event in your probability space. To put it less formally: for each of your possible outcomes, you have an associated value of $X$, and $X \le x$ is the event (set of outcomes) corresponding to values of $X$ that are less than $x$.
If we change $X \le x$ to $cX \le cx$ (for $c> 0$) or $X+c \le x+c$ or anything that describes the same values of $X$, we have a different description of the same event. There are no outcomes which make $X\le x$ occur but don't make $cX \le cx$ occur, or vice versa. So its probability will be the same.
We can always do anything to the inside of $\Pr[X \le x]$ that we could do and undo to an inequality. We are looking for the two-way implication $X \le x \iff X+c \le x+c$. As long as that holds, we're good.
Often, we can actually do more manipulations with the inside of a probability, that are only valid because of our specific experiment. For example, if we flip lots of coins, and $X$ is the number that land heads, then $X \le 2.5$ and $X \le 2$ are different descriptions of the same event. Even though they are not equivalent statements, there are no outcomes for which one event occurs, and the other does not.
A: Answer to both questions is yes. I'll come at this from a standpoint of not knowing anything about measure theory. For the first question, note that the area under the graph of a PDF must be $1.$ So we ask ourselves, what does the graph of the PDF of $2X$ look like compared to the graph of $X$? Well, it is spread out more by a factor of two in the $x$-direction. But if we left is at that, then the area under the PDF for $2X$ would be $2$. Therefore, we must make the following adjustment: if the PDF for the random variable $X$ is $f(x)$ with domain $a\leq x \leq b,\ $ then the PDF for the random variable $\ 2X\ $ is $\frac{1}{2}f\left(\frac{1}{2}x\right)$ with domain $\ 2a\leq x \leq 2b.\ $ But $2$ was arbitrary here, and so we get the following more general result:

if the PDF for the random variable $X$ is $f(x)$ with domain
$-\infty\leq a\leq x \leq b\leq \infty,\ $  then for $c>0,\ $ the PDF
for the random variable $\ cX\ $ is
$\frac{1}{c}f\left(\frac{1}{c}x\right)$ with domain $\ ca\leq x \leq cb.\ $

The diagram below demonstrates for $c=2$.
$$$$

The area of the two green rectangles is the same: the area of the rectangle in $\frac{1}{2}f\left(\frac{1}{2}x\right),$ is twice as short and twice as fat as it's corresponding rectangle on the $f(x)$ diagram. Thus "Area = $1$" has been preserved.
Since area is the limit of the sum of rectangles, these "approximate rectangles" can be formalised by the Lebesgue/measure theory. In fact, this is probably the purpose of Lebesgue/measure theory: to formalise this integration stuff.
Now, going back to your original question, which was:

is $P(X \leq x_0) = P(c X \leq cx_0)$ always true?

If we use some less technical results (than measure theory) like the fundamental theorem of calculus with the reverse chain rule, we can see that:
$$P(X \leq x_0) = \int_a^{x_0} f(x) dx = F(x_0) - F(a),$$
and
$$ P(c X \leq cx_0) = \int_{ca}^{cx_0} \frac{1}{c}f\left(\frac{1}{c}x\right)dx = \left[\frac{1}{c}\times cF\left(\frac{1}{c}x\right)\right]_{ca}^{cx_0} = F(x_0) - F(a), $$
attaining the required result.
A similar way of thinking about shifting of graphs can be done to attain the affirmative result for the second question.
